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- Thread starter jonathanm111
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I do know how to take the dot product but if its a simple matter with y=6x^2 then I must be missing something here. do you take the derivative of y and put it in dy and then solve for x and do the same thing? and then the dot product?## dr=\hat{i} dx+\hat{j} dy ## . With ## y=6x^2 ## it should be a simple matter to evaluate ## \int_C F \cdot dr ## over the path. Do you know how to take a dot product?(This one is in two dimensions instead of three).

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Additional problem: You can see that the line ## y=6x ## also connects the two endpoints. Try computing ## \int F \cdot dr ## over this straight line path between the same two points. Do you get the same answer? ## \\ ## If not, the integral is clearly path dependent. And if the answer is yes, the integral could still be path dependent for some other paths. The way to prove path dependence or independence is to look at ## \nabla \times F ##.

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alright i got it to work, its 1179/2.

And you'd get a different answer so its path-dependent. so we could actually use this information for BAdditional problem: You can see that the line ## y=6x ## also connects the two endpoints. Try computing ## \int F \cdot dr ## over this straight line path between the same two points. Do you get the same answer? ## \\ ## If not, the integral is clearly path dependent. And if the answer is yes, the integral could still be path dependent for some other paths. The way to prove path dependence or independence is to look at ## \nabla \times F ##.

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last quick question: how did you know to substitute the y into the x-integral

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That was a logical step, because it then made the x integral completely a function of ## x ##. ## \\ ## You could also have substituted that into the ## y ## integral, (## y=6x^2 ## ==>> ## dy=12x \, dx ##), and then solved the y-integral as a dx integral, using x-limits. That would have made extra unnecessary work though.last quick question: how did you know to substitute the y into the x-integral

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